<h2>
Speed of motorboat is 36 km/hr and speed of current is 4 km/hr.</h2>
Explanation:
Let speed of motor boat be m and speed of current be c.
A motorboat traveling with a current can go 160 km in 4 hours.
Distance = 160 km
Time = 4 hours
Speed = m + c
We have
Distance = Speed x Time
160 = (m+c) x 4
m + c = 40 --------------------- eqn 1
Against the current it takes 5 hours to go the same distance.
Distance = 160 km
Time = 5 hours
Speed = m - c
We have
Distance = Speed x Time
160 = (m-c) x 5
m - c = 32 --------------------- eqn 2
eqn 1 + eqn 2
2m = 40 + 32
m = 36 km/hr
Substituting in eqn 1
36 + c = 40
c = 4 km/hr
Speed of motorboat is 36 km/hr and speed of current is 4 km/hr.
Answer:
i think it might be D or C
Surveys are considered the most reliable way to gather data
Answer:=14,160 kJ
Explanation: Let m1 and m2 be the initial and final amounts of mass within the tank, respectively. The steam properties are listed in the table below
Specific Internal SpecificTemp Pressure Volume Energy Enthalpy Quality Phase
C MPa m^3/kg kJ/kg kJ/kg
1 260 4.689 0.02993 2158 2298 0.7 Liquid Vapor Mixture
2 260 4.689 0.0422 2599 2797 1 Saturated Vapor
The mass initially contained in the tank is m1 = V/v1
m1 =0.85 m^3 /0.02993 m^3 /kg
= 28.4 kg
The mass finally contained in the tank is
m2 =V2/v
= 0.85 m^3 /0.0422 m^3 /kg
= 20.14 kg
The heat transfer is then
Qcv = m2u2 − m1u1 − he(m2 − m1)
Qcv = (20.14)(2599) − (28.4)(2158) − (2797)(20.14 − 28.4) = 14,160 kJ
